What is the integral from 0 to r of br^3+cr^2 ?

The integral from 0 to r of br^3+cr^2 is (br^4)/4+(cr^3)/3

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integral from 0 to r of br^3+cr^2

\int_{0}^{r} b r^{3} + c r^{2} d r

\frac{b r ^{4}}{4} + \frac{c r ^{3}}{3}

Solution steps

\int_{0}^{r} b r^{3} + c r^{2} d r

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{r} b r^{3} d r + \int_{0}^{r} c r^{2} d r

\int_{0}^{r} b r^{3} d r = \frac{b r ^{4}}{4}

\int_{0}^{r} c r^{2} d r = \frac{c r ^{3}}{3}

= \frac{b r ^{4}}{4} + \frac{c r ^{3}}{3}

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What is the integral from 0 to r of br^3+cr^2 ?
The integral from 0 to r of br^3+cr^2 is (br^4)/4+(cr^3)/3